a: \(\lim\limits_{x\rightarrow2}\dfrac{1-\sqrt{x^2+3}}{-x^2+3x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+3}-1}{x^2-3x+2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{2^2+3}-1}{2^2-3\cdot2+2}\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\sqrt{2^2+3}-1=\sqrt{7}-1>0\\\lim\limits_{x\rightarrow2}2^2-3\cdot2+2=0\end{matrix}\right.\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{4x-1}+3}{x^2-4}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{4x-1-9}{\sqrt{4x-1}-3}\cdot\dfrac{1}{x^2-4}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}\cdot\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}=\dfrac{4\cdot2-10}{\sqrt{4\cdot2-1}-3}=\dfrac{-2}{\sqrt{7}-3}>0\\\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{1}{\left(2+2\right)\cdot\left(2-2\right)}=+\infty\end{matrix}\right.\)
